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Unformatted text preview: 382 8. MODULES If S Â M is any subset, we deﬁne \ annihilator of S to be ann.S / D
the
fr 2 R W rx D 0 for all x 2 S g D
ann.x/: Note that ann.S / is an
x 2S ideal, and ann.S / D ann.RS /, the annihilator of the submodule generated
by S . See Exercise 8.5.1
Consider a torsion module M over R. If S is a ﬁnite subset of M then
ann.S/ D ann.RS / is a nonzero ideal of R; in fact, if S D fx1 ; : : : ; xn g
and for each i , ri is a nonzero element of R such that ri xi D 0, then
Q
i ri is a nonzero element of ann.S /. If M is a ﬁnitely generated torsion
module, it follows that ann.M / is a nonzero ideal of R.
For the remainder of this section, R again denotes a principal idea
domain and M denotes a (nonzero) ﬁnitely generated module over R.
For x 2 Mtor , any generator of the ideal ann.x/ is called a period of
x . If a 2 R is a period of x 2 M , then Rx Š R=ann.x/ D R=.a/.
According to Lemma 8.4.5, any submodule of M is ﬁnitely generated.
If A is a torsion submodule of M , any generator of ann.A/ is a called a
period of A.
The period of an element x , or of a submodule A, is not unique, but
any two periods of x (or of A) are associates.
Lemma 8.5.5. Let M be a ﬁnitely generated module over a principal ideal
domain R.
(a) If M D A ˚ B , where A is a torsion submodule, and B is free,
then A D Mtor .
(b) M has a direct sum decomposition M D Mtor ˚ B , where B
is free. The rank of B in any such decomposition is uniquely
determined.
(c) M is a free module if, and only if, M is torsion free. Proof. We leave part (a) as an exercise. See Exercise 8.5.3. According to
the existence part of Theorem 8.5.2, M has a direct sum decomposition
M D A ˚ B , where A is a torsion submodule, and B is free. By part (a),
A D Mtor . Consequently, B Š M=Mtor , so the rank of B is determined.
This proves part (b).
For part (c), note that any free module is torsion free. On the other
hand, if M is torsion free, then by the decomposition of part (b), M is
free.
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 Fall '08
 EVERAGE
 Algebra

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